Problem: Which of the definite integrals is equivalent to the following limit? $ \lim_{n\to\infty} \sum_{i=1}^n \sin\left(\dfrac\pi4+\dfrac{\pi i}{2n}\right)\cdot\dfrac{\pi}{2n}$ Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{0}^{\pi /2} \sin x\,dx$ (Choice B) B $ \int_0^{\pi /4} \sin x\,dx$ (Choice C) C $ \int_{\pi /4}^{\pi} \sin x\,dx$ (Choice D) D $ \int_{\pi /4}^{3\pi /4} \sin x\,dx$
The value of a definite integral is the limit of its Riemann sums as the number of terms tends to infinity. The given summation $ \sum_{i=1}^n \sin\left(\dfrac\pi4+\dfrac{\pi i}{2n}\right)\cdot\dfrac{\pi}{2n}$ looks like a right Riemann sum with $n$ subintervals of equal width. If each subinterval has width $\Delta x$, what is the right Riemann sum for the following definite integral? $ \int_a^b f(x) \,dx$ The right Riemann sum for the definite integral is $ \sum_{i=1}^n f(a+i\Delta x)\cdot\Delta x$. What does the sum become when we express $\Delta x$ in terms of $a$, $b$, and $n$ ? Dividing the interval $[a,b]$ into $n$ subintervals of equal width yields a common width of $\Delta x=\dfrac{b-a}n\,$. This lets us express the right Riemann sum as $ \sum_{i=1}^n f\left(a+i\cdot\dfrac{b-a}n\right)\cdot\dfrac{b-a}n$. Let's rewrite the given summation as $ \sum_{i=1}^n \sin\left(\dfrac\pi4 + i\cdot\dfrac{\pi/2}n\right) \cdot\dfrac{\pi/2}n\,$. If the given summation is a right Riemann sum, what are $a$, $b$, and $f$ ? Equating the width $\Delta x$ of each subinterval in the two sums yields $\dfrac{b-a}n=\dfrac{\pi/2}n\,$. Thus, the interval $[a,b]$ has width $b-a=\pi/2$. In all the answer choices, $f(x)=\sin x$. Therefore, $a=\pi/4$ and $b=a+\pi/2=3\pi/4$. These values produce the definite integral $ \int_{\pi/4}^{3\pi/4} \sin x \,dx$. The correct answer is $ \lim_{n\to\infty} \sum_{i=1}^n \sin\left(\dfrac\pi4+\dfrac{\pi i}{2n}\right)\cdot\dfrac\pi{2n}=\int_{\pi /4}^{3\pi/4} \sin x \,dx$.